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NCERT Class 10 Mathematics Chapter 13 Statistics
Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 10 Mathematics Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 10 Mathematics Chapter 13 Statistics Solutions for All Subject, You can practice these here.
Statistics
Chapter – 13
| Exercise 13.1 |
1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
| Number of plants | 0 – 2 | 2 – 4 | 4 – 6 | 6 – 8 | 8 – 10 | 10 – 12 | 12 – 14 |
| Number of houses | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
Which method did you use for finding the mean, and why?
Ans: To find the class mark (xi) for each interval, the following relation is used.
Class mark 𝒙1 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕+𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/ 2
| Number of plants | Number of houses (fi) | xi | fixi |
| 0 − 2 | 1 | 1 | 1 × 1 = 1 |
| 2 − 4 | 2 | 3 | 2 × 3 = 6 |
| 4 − 6 | 1 | 5 | 1 × 5 = 5 |
| 6 − 8 | 5 | 7 | 5 × 7 = 35 |
| 8 − 10 | 6 | 9 | 6 × 9 = 54 |
| 10 − 12 | 2 | 1 | 2 ×11 = 22 |
| 12 − 14 | 3 | 13 | 3 × 13 = 39 |
| Total | 20 | 162 |
From the table, it can be observed that
Therefore, mean number of plants per house is 8.1.
2. Consider the following distribution of daily wages of 50 worker of a factory.
| Daily wages (in Rs) | 100 − 120 | 120 − 140 | 140 −1 60 | 160 − 180 | 180 − 200 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Ans: To find the class mark for each interval, the following relation is used.
X1 = upper limit + Lower Limit// 2
Class size (h) of this data = 20
Taking 150 as assured mean (a) , d1, u1 and f1u1 can be calculate as follows
| Daily wages (in Rs) | Number of workers (fi) | xi | di = xi − 150 | 𝒖𝒊 = 𝒅𝒊 /𝟐0 | fiui |
| 100 −120 | 12 | 110 | − 40 | -2 | -24 |
| 120 − 140 | 14 | 130 | − 20 | -1 | 14 |
| 140 − 160 | 8 | 150 | 0 | 0 | 0 |
| 160 −180 | 6 | 170 | 20 | 1 | 6 |
| 180 − 200 | 10 | 190 | 40 | 2 | 20 |
| Total | 50 | -12 |
From the table, it can be observed that
Therefore, the mean daily wage of the workers of the factory is Rs 145.20
3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency f.
| Daily pocket allowance (in Rs) | 11−13 | 13−15 | 15−17 | 17−19 | 19−21 | 21−23 | 23−25 |
| Number of workers | 7 | 6 | 9 | 13 | f | 5 | 4 |
Ans: To find the class mark (xi) for each interval, the following relation is used:
X1 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/2
Given that, mean pocket allowance,
Taking 18 as assured mean (a), di and f1d1 are calculated as follows.
| Daily pocket allowance (in Rs) | Number of children fi | Class mark xi | di = xi − 18 | fidi |
| 11 −13 | 7 | 12 | -6 | -42 |
| 13 − 15 | 6 | 14 | -4 | -24 |
| 15 − 17 | 9 | 16 | -2 | -18 |
| 17 −19 | 13 | 18 | 0 | 0 |
| 19 − 21 | f | 20 | 2 | 2f |
| 21 − 23 | 5 | 22 | 4 | 20 |
| 23 − 25 4 | 4 | 24 | 6 | 24 |
| Total |  | 2f-40 |
Hence, the missing frequency, f, is 20.
4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.
| Number of heart beats per minute | 65−68 | 68−71 | 71−74 | 74−77 | 77−80 | 80−83 | 83−8 |
| Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Ans: To find the class mark of each interval (xi), the following relation is used.
X1 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊 / 2
Class size, h, of this data = 3
Taking 75.5 as assumed mean (a) assumed (a) ui, f1u1 are calculated as follows.
| Number of heart beats per minute | Number of women fi | xi | di = xi − 75.5 | 𝒖𝒊 = 𝒅𝒊/3 | fiui |
| 65 − 68 | 2 | 66.5 | − 9 | -3 | -6 |
| 68 − 71 | 4 | 69.5 | − 6 | -2 | -8 |
| 71 − 74 | 3 | 69.5 | − 3 | -1 | -3 |
| 74 − 77 | 8 | 75.5 | 0 | 0 | 0 |
| 77 − 80 | 7 | 78.5 | 3 | 3 | 7 |
| 80 − 83 | 4 | 81.5 6 | 6 | 6 | 8 |
| 83 − 86 | 2 | 84.5 | 9 | 9 | 6 |
| Total | 30 | 4 |
Therefore, mean hear beats per minute for these women are 75.9 beats per minute.
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
| Number of mangoes | 50 − 52 | 53 − 55 | 56−58 | 59−61 | 62−64 |
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Ans:
| Number of mangoes | Number of boxes fi |
| 50 − 52 | 15 |
| 53 − 55 | 110 |
| 56 − 58 | 135 |
| 59 − 61 | 115 |
| 62 − 64 | 25 |
It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore, 1/2 has to be added to the upper class limit.
and 1/2 has to be subtracted from the lower class limit of each interval.
Class mark (xi) can be obtained by using the following relation.
𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐
Class size (h) of this data = 3
Taking 57 as assumed mean (a), di, ui, f1u1are calculated as follows
| Class interval | fi | xi | di = xi − 57 | 𝒖𝒊 = 𝒅i/3 | f1xu1 |
| 49.5 − 52.5 | 15 | 51 | -6 | -2 | -30 |
| 52.5 − 55.5 | 110 | 54 | -3 | -1 | -110 |
| 55.5 − 58.5 | 135 | 57 | 0 | 0 | 0 |
| 58.5 − 61.5 | 115 | 60 | 3 | 1 | 115 |
| 61.5 − 64.5 | 25 | 63 | 6 | 2 | 50 |
| Total 4 | 400 |
Mean number of mangoes kept in a packing box is 57.19.
Step deviation method is used here as the values of fi, di are big and also, there is a common multiple between all di.
6. The table below shows the daily expenditure on food of 25 households in a locality.
| Daily expenditure (in Rs) | 100−150 | 150−200 | 200−250 | 250−300 | 300−350 |
| Number of households | 4 | 5 | 12 | 2 | 3 |
Find the mean daily expenditure on food by a suitable method.
Ans: To find the class mark (xi) for each interval, the following relation is used.
𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐
Class size = 50
Taking 225 as assumed mean (a), di, ui, f1u1are calculated as follows.
| Daily expenditure (in Rs) | f1 | xi | di = xi − 225 | 𝒖𝒊 = 𝒅𝒊 50 f | f1u1 |
| 100 − 150 | 4 | 125 | -100 | -2 | -8 |
| 100 − 150 | 5 | 175 | -50 | -1 | -5 |
| 200 − 250 | 12 | 225 | 0 | 0 | 0 |
| 250 − 300 | 2 | 275 | 50 | 1 | 2 |
| 300 − 350 | 2 | 325 | 100 | 2 | 4 |
| Total | 25 |
Therefore, mean daily expenditure on food is Rs 211.
7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
| concentration of SO2 (in ppm) | Frequency |
| 0.00 − 0.04 | 4 |
| 0.04 − 0.08 | 9 |
| 0.08 − 0.12 | 9 |
| 0.12 − 0.16 | 2 |
| 0.16 − 0.20 | 4 |
| 0.20 − 0.24 | 2 |
Ans: To find the class marks for each interval, the following relation is used.
𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊/2
lass size of this data = 0.04
Taking 0.14 as assumed mean (a), di, ui, f1u1are calculated as follows.
| Concentration of SO2 (in ppm) | Frequency f | Class mark xi | di = xi − 0.14 | 𝒖𝒊 = 𝒅𝒊 0.04 | fiui |
| 0.00 − 0.04 | 4 | 0.02 | -0.12 | − 3 | − 12 |
| 0.04 − 0.08 | 9 | 0.06 | -0.08 | − 2 | − 18 |
| 0.08 − 0.12 | 9 | 0.10 | -0.04 | − 1 | − 9 |
| 0.12 − 0.16 | 2 | 0.14 | 0 | 0 | 0 |
| 0.16 − 0.20 | 4 | 0.18 | 1 | 1 | 4 |
| 0.20 − 0.24 | 2 | 0.22 | 2 | 2 | 4 |
| Total | 30 | −31 |
Therefore, mean concentration of SO2 in the air is 0.099 ppm.
8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
| Number of days | 0−6 | 6−10 | 10−14 | 14−20 | 20−28 | 20−28 | 38−40 |
| Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Ans: To find the class mark of each interval, the following relation is used.
𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊/2
Taking 17 as assumed mean (a), di and f1d1 are calculated as follows.
| Number of days | Number of students | xi | di = xi − 17 | fid |
| 0 − 6 | 11 | 3 | − 14 | − 154 |
| 6 − 10 | 10 | 8 | − 9 | − 90 |
| 10 − 14 | 7 | 12 | − 5 | − 35 |
| 14 − 20 | 4 | 4 | 0 | 0 |
| 20 − 28 | 4 | 24 | 7 | 28 |
| 28 − 38 | 3 | 33 | 16 | 48 |
| 38 − 40 | 1 | 39 | 22 | 22 |
| Total | 40 | − 181 |
From the table, we obtain
Therefore, the mean number of days is 12.48 days for which a student was absent.
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
| Literacy rate (in %) | 45 − 55 | 55 − 65 | 65 − 75 | 75 − 85 | 85 − 95 |
| Number of cities | 3 | 10 | 11 | 8 | 3 |
Ans: To find the class marks, the following relation is used.
𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐
Class size (h) for this data = 10
Taking 70 as assumed mean (a), di, ui, and fiui are calculated as follows.
| Literacy rate (in %) | Number of cities fi | x1 | di = x1 − 70 | 𝒖𝒊 = 𝒅𝒊 1 | f1 u1 |
| 45 − 55 | 3 | 50 | -20 | -2 | -6 |
| 55 − 65 | 10 | 60 | -10 | -10 | -10 |
| 65 − 75 | 11 | 70 | -0 | 0 | 0 |
| 75 − 85 | 8 | 80 | 10 | 1 | 8 |
| 85 − 95 | 3 | 90 | 20 | 2 | 6 |
| Total | 35 | − 2 |
From the table, we obtain
Therefore, mean literacy rate is 69.43%.
| Exercise 13.2 |
1. The following table shows the ages of the patients admitted in a hospital during a year:
| Age (in years) | 5−15 | 15 − 25 | 25−35 | 35−45 | 45−55 | 55−65 |
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Ans: To find the class marks (xi), the following relation is used.
𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐
Taking 30 as assumed mean (a), di and f1d1are calculated as follows.
| Age (in years) | Number of patients fi | Class mark xi | di = xi − 30 | fidi |
| 5 − 15 | 6 | 10 | -20 | -120 |
| 15 − 25 | 11 | 20 | – 10 | -110 |
| 25 − 35 | 21 | 30 | 0 | 0 |
| 35 − 45 | 23 | 40 | 10 | 230 |
| 45 − 55 | 14 | 50 | 20 | 280 |
| 55 − 65 | 5 | 60 | 30 | 150 |
| Total | 80 | 430 |
Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years.
It can be observed that the maximum class frequency is 23 belonging to class interval 35 − 45.
Modal class = 35 − 45
Lower limit (l) of modal class = 35
Frequency (f1) of modal class = 23
Class size (h) = 10
Frequency (f0) of class preceding the modal class = 21
Frequency (f2) of class succeeding the modal class = 14
2. Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.
| Lifetimes (in hours) | 0 − 20 | 20−40 | 40−60 | 60−80 | 80−100 | 100−120 |
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.
| Expenditure (in Rs) | Number of families |
| 1000 − 1500 | 24 |
| 1500 − 2000 | 40 |
| 2000 − 2500 | 33 |
| 2500 − 3000 | 28 |
| 3000 − 3500 | 30 |
| 3500 − 4000 | 22 |
| 4000 − 4500 | 16 |
| 4500 − 5000 | 7 |
4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
| Number of students per teacher | Number of states/U.T |
| 15 − 20 | 3 |
| 20 − 25 | 8 |
| 25 − 30 | 9 |
| 30 − 35 | 10 |
| 35 − 40 | 3 |
| 40 − 45 | 0 |
| 45 − 50 | 0 |
| 50 − 55 | 2 |
Ans: It can be observed from the given data that the maximum class frequency is 10
belonging to class interval 30 − 35.
Therefore, modal class = 30 − 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f1) of modal class = 10
Frequency (f0) of class preceding modal class = 9
Frequency (f2) of class succeeding modal class = 3
Mode = 30.6
It represents that most of the states/U.T have a teacher-student ratio as 30.6.
To find the class marks, the following relation is used.
𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/2
Taking 32.5 as assumed mean (a), di, ui, and fiui are calculated as follows.
| Number of students per teacher | Number of states/U.T (fi) | xi | di = xi − 32.5 | 𝒖𝒊 = 𝒅𝒊/5 | f1u1 |
| 15 − 20 | 3 | 17.5 | − 15 | -3 | − 9 |
| 20 − 25 | 8 | 22.5 | − 10 | -2 | −16 |
| 25 − 30 | 9 | 27.5 | − 5 | -1 | − 9 |
| 30 − 35 | 10 | 32.5 | 0 | 0 | 0 |
| 35 − 40 | 3 | 37.5 | 5 | 1 | 3 |
| 40 − 45 | 0 | 42.5 | 10 | 2 | 0 |
| 45 − 50 | 0 | 47.5 | 15 | 3 | 0 |
| 50 − 55 | 2 | 52.5 | 20 | 4 | 8 |
| Total | 35 | −23 |
Therefore, mean of the data is 29.2.
It represents that on an average, teacher−student ratio was 29.2.
5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
| Runs scored | Number of batsmen |
| 3000 − 4000 | 4 |
| 4000 − 5000 | 18 |
| 5000 − 6000 | 9 |
| 6000 − 7000 | 7 |
| 7000 − 8000 | 6 |
| 8000 − 9000 | 3 |
| 9000 − 10000 | 1 |
| 10000 − 11000 | 1 |
Find the mode of the data.
Ans: From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000−5000. Therefore, modal class = 4000−5000
Lower limit (l) of modal class = 4000
Frequency (f1) of modal class = 18
Frequency (f0) of class preceding modal class = 4
Frequency (f2) of class succeeding modal class = 9
Class size (h) = 1000
Therefore, mode of the given data is 4608.7 runs
6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
| Number of cars | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
| Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Ans:
| Exercise 13.3 |
1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
| Monthly consumption (in units) | Number of consumers |
| 65 − 85 | 4 |
| 85 − 105 | 5 |
| 105 − 125 | 13 |
| 125 − 145 | 20 |
| 145 − 165 | 14 |
| 165 − 185 | 8 |
| 185 − 205 | 4 |
Ans: To find the class marks, the following relation is used.
𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐
Taking 135 as assumed mean (a), di, ui, fiui are calculated according to step deviation method as follows.
| Monthly consumption (in units) | Number of consumers (f i) | xi class mark | di= xi− 135 | 𝒖𝒊 = 𝒅𝒊 𝟐𝟎 | fiui |
| 65 − 85 | 4 | 75 | − 60 | − 3 | −12 |
| 85 − 105 | 5 | 95 | − 40 | − 2 | − 10 |
| 105 − 125 1 | 13 | 115 | − 40 | − 1 | − 13 |
| 125 − 145 | 20 | 135 | 0 | 0 | 0 |
| 145 − 165 | 14 | 155 | 20 | 1 | 14 |
| 165 − 185 | 8 | 175 | 40 | 2 | 16 |
| 185 − 205 | 4 | 195 | 60 | 3 | 12 |
| Total | 68 | 7 |
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.
Modal class = 125 − 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 13
Frequency (f2) of class succeeding the modal class = 14
(ii) To find the median of the given data, cumulative frequency is calculated as follows.
| Monthly consumption (in units) | Number of consumers | Cumulative frequency |
| 65 − 85 | 4 | 4 |
| 85 − 105 | 5 | 4 + 5 = 9 |
| 105 − 125 | 13 | 9 + 13 = 22 |
| 125 − 145 | 20 | 22 + 20 = 42 |
| 145 − 165 | 14 | 42 + 14 = 56 |
| 165 − 185 | 8 | 56 + 8 = 64 |
| 185 − 205 | 4 | 64 + 4 = 68 |
From the table, we obtain n = 68 Cumulative frequency (cf) just greater than n/2
(i.e,68/2=34) is 42 belonging to interval 125 − 145.
Therefore, median class = 125 − 145
Lower limit (l) of median class = 125
Class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05
Respectively.
The three measures are approximately the same in this case.
2. If the median of the distribution is given below is 28.5, find the values of x and y.
| Class interval: 0-10 | 10 – 20 | 20 – 30 | 30 -40 | 40-50 | 50-60 | Total |
| Frequency: 5 | x | 20 | 15 | y | 5 | 60 |
Ans: The cumulative frequency for the given data is calculated as follows.
| 50 − 60 | Frequency | Cumulative frequency |
| 0 − 10 | 5 | 5 |
| 10 − 20 | x | 5+ x |
| 20 − 30 | 20 | 25 + x |
| 30 − 40 | 15 | 40 + x |
| 40 − 50 | y | 40+ x + y |
| 50 − 60 | 5 | 45 + x + y |
| Total (n) | 60 |
From the table, it can be observed that n = 60
45 + x + y = 60 or x + y = 15 ..……………….(1)
Median of the data is given as 28.5 which lies in interval 20 − 30.
Therefore, median class = 20 − 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
From equation (1),
8 + y = 15 y = 7
Hence, the values of x and y are 8 and 7 respectively
3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
| Age (in years) | Number of policy holders |
| Below 20 | 2 |
| Below 25 | 6 |
| Below 30 | 24 |
| Below 35 | 45 |
| Below 40 | 78 |
| Below 45 | 89 |
| Below 50 | 92 |
| Below 55 | 98 |
| Below 60 | 100 |
Ans: Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.
| Age (in years) | Number of policy holders (fi) | Cumulative frequency (cf) |
| 18 − 20 | 2 | 2 |
| 20 − 25 | 6 − 2 = 4 | 6 |
| 25 − 30 | 24 − 6 = 18 | 24 |
| 30 − 35 | 45 − 24 = 21 | 45 |
| 35 − 40 | 78 − 45 = 33 | 78 |
| 40 − 45 | 89 − 78 = 11 | 89 |
| 45 − 50 | 92 − 89 = 3 | 92 |
| 50 − 55 | 98 − 92 = 6 | 98 |
| 55 − 60 | 100 − 98 = 2 | 100 |
| Total (n) |
From the table, it can be observed that n = 100.
Cumulative frequency (cf) just greater than is 78, belonging to
interval 35 − 40
Therefore, median class = 35 − 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45
Therefore, the median age is 35.76 years.
4. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
| Length (in mm) | Number or leaves f |
| 118 − 126 | 3 |
| 127 − 135 | 5 |
| 136 − 144 | 9 |
| 145 − 153 | 12 |
| 154 − 162 | 5 |
| 163 − 171 | 4 |
| 172 − 180 | 2 |
Find the median length of the leaves.
Ans:
| Length (in mm) | Number or leaves f | Cumulative frequency |
| 117.5 − 126.5 | 3 | 3 |
| 126.5 − 135.5 | 5 | 3 + 5 = 8 |
| 135.5 − 144.5 | 9 | 8 + 9 = 17 |
| 144.5 − 153.5 | 12 | 17 + 12 = 29 |
| 153.5 − 162.5 | 5 | 29 + 5 = 34 |
| 162.5 − 171.5 | 4 | 34 + 4 = 38 |
| 171.5 − 180.5 | 2 | 38 + 2 = 40 |
From the table, it can be observed that the cumulative frequency just greater than
is 29, belonging to class interval 144.5 − 153.5.
Median class = 144.5 − 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17
Therefore, median length of leaves is 146.75 mm.
5. Find the following table gives the distribution of the life time of 400 neon lamps:
| Life time (in hours) | Number of lamps |
| 1500 − 2000 | 14 |
| 2000 − 2500 | 56 |
| 2500 − 3000 | 60 |
| 3000 − 3500 | 86 |
| 3500 − 4000 | 74 |
| 4000 − 4500 | 62 |
| 4500 − 5000 | 48 |
Ans: The cumulative frequencies with their respective class intervals are as follows.
| Life tim | Number of lamps (fi) | Cumulative frequency |
| 1500 − 2000 | 14 | 14 |
| 2000 − 2500 | 56 | 14 + 56 = 70 |
| 2500 − 3000 | 60 | 70 + 60 = 130 |
| 3000 − 3500 | 86 | 130 + 86 = 216 |
| 3500 − 4000 | 74 | 216 + 74 = 290 |
| 4000 − 4500 | 62 | 290 + 62 = 352 |
| 4500 − 5000 | 48 | 352 + 48 = 400 |
| Total (n) | 400 |
It can be observed that the cumulative frequency just greater than n/2 (i.e 400/2 = 200)
Median class = 3000 − 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
= 3406.976
Therefore, median life time of lamps is 3406.98 hours.
6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
| Number of letters | 1−4 | 4−7 | 7−10 | 10−13 | 13−16 | 16−19 |
| Number of surnames | 6 | 30 | 40 | 6 | 4 | 4 |
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Ans: The cumulative frequencies with their respective class intervals are as follows:
| Number of letters | Frequency (fi) | Cumulative frequency |
| 1 − 4 | 6 | 6 |
| 4 − 7 | 30 | 30 + 6 = 36 |
| 7 − 10 | 40 | 36 + 40 = 76 |
| 10 − 13 | 16 | 76 + 16 = 92 |
| 13 − 16 | 4 | 92 + 4 = 96 |
| 16 − 19 | 4 | 96 + 4 = 100 |
| Total (n) | 100 |
It can be observed that the cumulative frequency just greater than n/2 (i .e , 100/2 = 50)
is 76, belonging to class interval 7 − 10
Median class = 7 − 10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
= 8.05
To find the class marks of the given class intervals, the following relation is used.
𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐
Taking 11.5 as assumed mean (a), di, ui, and fiui are calculated according to step deviation method as follows.
(ii) Taking 11.5 as assumed mean (a), di, ui, and fiui are calculated according to step deviation method as follows.
Ans:
| Number of letters | Number of surnames fi | fi xi | di = xi− 11.5 | 𝒖𝒊 = 𝒅𝒊 𝟑 | fiui |
| 1 − 4 | 6 | 2.5 | − 9 | − 3 | − 18 |
| 4 − 7 | 30 | 5.5 | − 6 | − 2 | − 60 |
| 7 − 10 | 40 | 8.5 | − 3 | − 1 | − 40 |
| 10 − 13 | 16 | 11.5 | 0 | 0 | 0 |
| 13 − 16 | 4 | 14.5 | 3 | 1 | 4 |
| 16 − 19 | 4 | 17.5 | 6 | 2 | 8 |
| Total | 100 |
From the table, we obtain
Mean,
= 11.5 − 3.18 = 8.32
The data in the given table can be written as
| Number of letters | Frequency (fi) |
| 1 − 4 | 6 |
| 4 − 7 | 30 |
| 7 − 10 | 40 |
| 13 − 16 | 16 |
| 13 − 16 | 4 |
| 16 − 19 | 4 |
| 100 |
From the table, it can be observed that the maximum class frequency is 40
belonging to class interval 7 − 10. Modal class = 7 − 10
Lower limit (l) of modal class = 7
Class size (h) = 3
Frequency (f1) of modal class = 40
Frequency (f0) of class preceding the modal class = 30
Frequency (f2) of class succeeding the modal class = 16
Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88.
7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
| Weight (in kg) | 40−45 | 45−50 | 50−55 | 55−60 | 60−65 | 65−70 | 70−75 |
| Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |
Ans: The cumulative frequencies with their respective class intervals are as follows:
| Weight (in kg) | Frequency (fi) | Cumulative frequency |
| 40 − 45 | 2 | 2 |
| 45 − 50 | 3 | 2 + 3 = 5 |
| 50 − 55 | 8 | 5 + 8 = 13 |
| 55 − 60 | 6 | 13 + 6 = 19 |
| 60 − 65 | 6 | 19 + 6 = 25 |
| 65 − 70 | 3 | 25 + 3 = 28 |
| 70 − 75 | 2 | 28 + 2 = 30 |
| Total (n) | 2 |
Cumulative frequency just greater than is 19, belonging to class
interval 55 − 60.
Median class = 55 − 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
= 56.67
Therefore, the median weight is 56.67 kg.